The double-angle formula $\sin2x=2\sin x\cos x$ turns the scary-looking integral $$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{2^k}$$ into fun once you realize $\prod_k\cos\frac{z}{2^k}=\frac{\sin z}z$, because then it's well-known that $\int_0^{\infty}\frac{\sin z}zdz=\frac{\pi}2$.

I've found the following variant intriguing and curious.

Question. Is this valid? If not, what is the value of the integral? $$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{k}=\frac{\pi}4.$$

In case such is known, please provide me with a reference. Thanks.

The double-angle formula, $\sin2x=2\sin x\cos x$, turns the scary-looking integral $$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{2^k}$$ into fun once you realize $\prod_k\cos\frac{z}{2^k}=\frac{\sin z}z$, because then it's well-known that $\int_0^{\infty}\frac{\sin z}zdz=\frac{\pi}2$.

I've found the following variant intriguing and curious.

Question. Is this valid? If not, what is the value of the integral? $$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{k}=\frac{\pi}4.$$

In case such is known, what is a reference?